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Chemistry, The Central Science , 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations. John D. Bookstaver St. Charles Community College Cottleville, MO. Antoine Lavoisier.

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chapter 3 stoichiometry calculations with chemical formulas and equations

Chemistry, The Central Science, 11th edition

Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

Chapter 3Stoichiometry:Calculations with Chemical Formulas and Equations

John D. Bookstaver

St. Charles Community College

Cottleville, MO

© 2009, Prentice-Hall, Inc.

antoine lavoisier
Antoine Lavoisier

© 2009, Prentice-Hall, Inc.

law of conservation of mass
Law of Conservation of Mass

“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.”

--Antoine Lavoisier, 1789

© 2009, Prentice-Hall, Inc.

chemical equations
Chemical Equations

Chemical equations are concise representations of chemical reactions.

© 2009, Prentice-Hall, Inc.

anatomy of a chemical equation7
Reactants appear on the left side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Anatomy of a Chemical Equation

© 2009, Prentice-Hall, Inc.

anatomy of a chemical equation8
Products appear on the right side of the equation.

CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(g)

Anatomy of a Chemical Equation

© 2009, Prentice-Hall, Inc.

anatomy of a chemical equation9
The states of the reactants and products are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(g)

Anatomy of a Chemical Equation

© 2009, Prentice-Hall, Inc.

anatomy of a chemical equation10
Coefficients are inserted to balance the equation.

CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(g)

Anatomy of a Chemical Equation

© 2009, Prentice-Hall, Inc.

subscripts and coefficients give different information
Subscripts and Coefficients Give Different Information
  • Subscripts tell the number of atoms of each element in a molecule.

© 2009, Prentice-Hall, Inc.

subscripts and coefficients give different information12
Subscripts and Coefficients Give Different Information
  • Subscripts tell the number of atoms of each element in a molecule
  • Coefficients tell the number of molecules.

© 2009, Prentice-Hall, Inc.

reaction types

Reaction Types

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combination reactions
Examples:

2 Mg (s) + O2 (g)  2 MgO (s)

N2 (g) + 3 H2 (g) 2 NH3 (g)

C3H6 (g) + Br2 (l)  C3H6Br2 (l)

In this type of reaction two or more substances react to form one product.

Combination Reactions

© 2009, Prentice-Hall, Inc.

decomposition reactions
Decomposition Reactions
  • In a decomposition one substance breaks down into two or more substances.
  • Examples:
    • CaCO3(s) CaO (s) + CO2(g)
    • 2 KClO3(s)  2 KCl (s) + O2 (g)
    • 2 NaN3(s) 2 Na (s) + 3 N2(g)

© 2009, Prentice-Hall, Inc.

combustion reactions
Examples:

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

These are generally rapid reactions that produce a flame.

Most often involve hydrocarbons reacting with oxygen in the air.

Combustion Reactions

© 2009, Prentice-Hall, Inc.

formula weights

Formula Weights

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formula weight fw
Formula Weight (FW)
  • A formula weight is the sum of the atomic weights for the atoms in a chemical formula.
  • So, the formula weight of calcium chloride, CaCl2, would be

Ca: 1(40.1 amu)

+ Cl: 2(35.5 amu)

111.1 amu

  • Formula weights are generally reported for ionic compounds.

© 2009, Prentice-Hall, Inc.

molecular weight mw

+ H: 6(1.0 amu)

Molecular Weight (MW)
  • A molecular weight is the sum of the atomic weights of the atoms in a molecule.
  • For the molecule ethane, C2H6, the molecular weight would be

C: 2(12.0 amu)

30.0 amu

© 2009, Prentice-Hall, Inc.

percent composition

(number of atoms)(atomic weight)

x 100

% element =

(FW of the compound)

Percent Composition

One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

© 2009, Prentice-Hall, Inc.

percent composition21

(2)(12.0 amu)

%C =

(30.0 amu)

x 100

24.0 amu

=

30.0 amu

Percent Composition

So the percentage of carbon in ethane is…

= 80.0%

© 2009, Prentice-Hall, Inc.

moles

Moles

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avogadro s number
Avogadro’s Number
  • 6.02 x 1023
  • 1 mole of 12C has a mass of 12 g.

© 2009, Prentice-Hall, Inc.

molar mass
Molar Mass
  • By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).
    • The molar mass of an element is the mass number for the element that we find on the periodic table.
    • The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

© 2009, Prentice-Hall, Inc.

using moles
Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

© 2009, Prentice-Hall, Inc.

mole relationships
Mole Relationships
  • One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.
  • One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.
  • Remember: National Mole Moment is at 6:02 on October 23rd each year!

© 2009, Prentice-Hall, Inc.

finding empirical formulas

Finding Empirical Formulas

© 2009, Prentice-Hall, Inc.

calculating empirical formulas
Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition.

© 2009, Prentice-Hall, Inc.

calculating empirical formulas29
Calculating Empirical Formulas

The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

© 2009, Prentice-Hall, Inc.

calculating empirical formulas30

Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol

12.01 g

1 mol

14.01 g

1 mol

16.00 g

1 mol

1.01 g

Calculating Empirical Formulas

© 2009, Prentice-Hall, Inc.

calculating empirical formulas31

C: = 7.005  7

H: = 6.984  7

N: = 1.000

O: = 2.001  2

5.09 mol

0.7288 mol

0.7288 mol

0.7288 mol

5.105 mol

0.7288 mol

1.458 mol

0.7288 mol

Calculating Empirical Formulas

Calculate the mole ratio by dividing by the smallest number of moles:

© 2009, Prentice-Hall, Inc.

calculating empirical formulas32
Calculating Empirical Formulas

These are the subscripts for the empirical formula:

C7H7NO2

© 2009, Prentice-Hall, Inc.

combustion analysis
Combustion Analysis
  • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.
    • C is determined from the mass of CO2 produced.
    • H is determined from the mass of H2O produced.
    • O is determined by difference after the C and H have been determined.

© 2009, Prentice-Hall, Inc.

elemental analyses
Elemental Analyses

Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.

© 2009, Prentice-Hall, Inc.

stoichiometric calculations
Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.

© 2009, Prentice-Hall, Inc.

stoichiometric calculations36
Stoichiometric Calculations

Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

© 2009, Prentice-Hall, Inc.

stoichiometric calculations37
Stoichiometric Calculations

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams.

© 2009, Prentice-Hall, Inc.

limiting reactants

Limiting Reactants

© 2009, Prentice-Hall, Inc.

how many cookies can i make
How Many Cookies Can I Make?
  • You can make cookies until you run out of one of the ingredients.
  • Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

© 2009, Prentice-Hall, Inc.

how many cookies can i make40
How Many Cookies Can I Make?
  • In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

© 2009, Prentice-Hall, Inc.

limiting reactants41
Limiting Reactants
  • The limiting reactant is the reactant present in the smallest stoichiometric amount.
    • In other words, it’s the reactant you’ll run out of first (in this case, the H2).

© 2009, Prentice-Hall, Inc.

limiting reactants42
Limiting Reactants

In the example below, the O2 would be the excess reagent.

© 2009, Prentice-Hall, Inc.

theoretical yield
Theoretical Yield
  • The theoretical yield is the maximum amount of product that can be made.
    • In other words it’s the amount of product possible as calculated through the stoichiometry problem.
  • This is different from the actual yield, which is the amount one actually produces and measures.

© 2009, Prentice-Hall, Inc.

percent yield

Actual Yield

Theoretical Yield

Percent Yield = x 100

Percent Yield

One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).

© 2009, Prentice-Hall, Inc.